I think, that you can not do it that brute way (and use rationals, not floats).

You "inner" integral is something like a momentum and has a formal solution,
J:= (1/12*erf(1/30*5^(1/2)*3^(1/2)*(6*x-z))+1/12)*z-
    1/6*5^(1/2)*3^(1/2)/Pi^(1/2)*exp(-1/60*(6*x-z)^2)

Note that for z=0 there is no solution. And I think you have to approximate that J
first instead of using fsolve or RootOf (and evalf), especially in tails, where
both erf and exp may be very flat.

For me the following works:

#eval(F(s), s = x - a) and then residue F(x-a) in x=0
eval(1/(2^s-1), s = x - 2*Pi*I*3/log(2));
series(%, x=0);
coeff(%, x, -1);

         1    -1                                   3  3      4
       ----- x   - 1/2 + 1/12 ln(2) x - 1/720 ln(2)  x  + O(x )
       ln(2)

                                  1
                                -----
                                ln(2)

A := convert(LommelS1(a, -1+a, z)/z^(a+1), hypergeom);
my:=1/4*1/a*hypergeom([1],[2, a+1],-1/4*z^2);

 'A-my';
  simplify(%, power) assuming 2 < a;
  convert(%, Sum, only=hypergeom, dummy=k);
  combine(%) assuming 0 < a, 0<z;
  simplify(%);
  #simplify(%) assuming 0 < a, 0<z;
  value(%);


                                  0

A := convert(LommelS1(a, -1+a, z)/z^(a+1), hypergeom);
my:=1/4*1/a*hypergeom([1],[2, a+1],-1/4*z^2);

 'A-my';
  simplify(%, power) assuming 2 < a;
  convert(%, Sum, only=hypergeom, dummy=k);
  combine(%) assuming 0 < a, 0<z;
  simplify(%);
  #simplify(%) assuming 0 < a, 0<z;
  value(%);


                                  0

That plot uses default settings for plotting and it is not
enough for values close to 0, due to numerical errors.

check:=proc(a,z)
Digits:=36;
evalf( (z^(-1+a)-2*2^(-1+a)*z^(-1+a)*
  hypergeom([],[a],-1/4*z^2)/(2^a))/(z^(a+1)) );
end proc;

plot(check(4.04,z), z=0 .. 1e-3);

That plot uses default settings for plotting and it is not
enough for values close to 0, due to numerical errors.

check:=proc(a,z)
Digits:=36;
evalf( (z^(-1+a)-2*2^(-1+a)*z^(-1+a)*
  hypergeom([],[a],-1/4*z^2)/(2^a))/(z^(a+1)) );
end proc;

plot(check(4.04,z), z=0 .. 1e-3);

The 2nd answer says: that is the symmetric group on 6 elements - hence p is not solvable by radicals

The 2nd answer says: that is the symmetric group on 6 elements - hence p is not solvable by radicals

I have not looked closer, but up to a shift ( = linear change of variables ) it is
like 1 / MillsRatio and I remember having numerical troubles with it once.

http://en.wikipedia.org/wiki/Mills_ratio,
http://en.wikipedia.org/wiki/Inverse_Mills_ratio

However Int(exp(-t^2/2)/erfc(1/2*t*2^(1/2)),t = -infinity .. infinity) will diverge (?).
# ..................^^

I have not looked closer, but up to a shift ( = linear change of variables ) it is
like 1 / MillsRatio and I remember having numerical troubles with it once.

http://en.wikipedia.org/wiki/Mills_ratio,
http://en.wikipedia.org/wiki/Inverse_Mills_ratio

However Int(exp(-t^2/2)/erfc(1/2*t*2^(1/2)),t = -infinity .. infinity) will diverge (?).
# ..................^^

I once wanted something like real modulo real*Integer, http://www.mapleprimes.com/forum/modulonotonlyintegers

For w mod ( w/(2*Pi/26) ) = w mod ( 13*w/Pi ) however I do not understand, what is desired:

If you read r modulo s as Reals / s*Integers then r modulo s means, that one computes r/s,
takes the nearest integer below and the difference is the 'modulo' (positives assumed).

Now r=w and s = w/(2*Pi/26) = 13*w/Pi ~ 4*w, so floor(r/s) = 0 and the difference is
r/s= w / ( 13*w/Pi ) = Pi/13.

Which does not depend on w.


Is there something wrong with the formula you use or do I have an error?

I once wanted something like real modulo real*Integer, http://www.mapleprimes.com/forum/modulonotonlyintegers

For w mod ( w/(2*Pi/26) ) = w mod ( 13*w/Pi ) however I do not understand, what is desired:

If you read r modulo s as Reals / s*Integers then r modulo s means, that one computes r/s,
takes the nearest integer below and the difference is the 'modulo' (positives assumed).

Now r=w and s = w/(2*Pi/26) = 13*w/Pi ~ 4*w, so floor(r/s) = 0 and the difference is
r/s= w / ( 13*w/Pi ) = Pi/13.

Which does not depend on w.


Is there something wrong with the formula you use or do I have an error?

The mathematical convention is: products (and thus powers) have a closer binding than summations and sign operators

It is a known bug in Excel (depending on versions?) try "=-1^2" in a worksheet giving -1
and "Debug.Print WorksheetFunction.Power(-1, 2)" in VBA, giving +1

The mathematical convention is: products (and thus powers) have a closer binding than summations and sign operators

It is a known bug in Excel (depending on versions?) try "=-1^2" in a worksheet giving -1
and "Debug.Print WorksheetFunction.Power(-1, 2)" in VBA, giving +1

Denote your according function by G. For large a one can 'ignore the 3' and then one
can solve the equation, roots are integers j or j +- arctan(sqrt(2)/Pi).

  G := a -> sin(a*Pi)*(4*a^2+3)^(1/2)*cos((4*a^2+3)^(1/2)*Pi)+
            sin((4*a^2+3)^(1/2)*Pi)*a*cos(a*Pi);

  x0:=0.5:
  for j from 1 to 10 do
    x0:=fsolve(G(a), a=x0+arctan(2^(1/2))/Pi):
    print(j, x0);
  end do:
                         1, 0.801320944698016
                         2, 1.09380384597767
                         3, 1.51655119238972
                         4, 1.90613590027121
                         5, 2.18336502326386
                         6, 2.58771773466317
                         7, 2.93753889015002
                         8, 3.21975661705503
                         9, 3.61805275567532
                         10, 3.95315218320903

This is with some 'luck', since a more sound way would try

  for j from 990 to 1000 do
    fsolve(G(a), a = j-arctan(sqrt(2)/Pi) .. j), 
    fsolve(G(a), a = j), 
    fsolve(G(a), a = j .. j+arctan(sqrt(2)/Pi));
  end do; 

For small j this would give repeated solution, so either one uses sets or writes
it in a proper way avoiding zeros already found (as others already showed).


PS: I looked at the equivalent task r1*sin(r1) = r2*sin(r2), r2 = r2=-r1-2*a*Pi
with r1 = Pi*(-a+(4*a^2+3)^(1/2)), but failed with analytic considerations.